Answer:
Buoyant force on the wood = 0.603 N
Buoyant force on the iron = 12.722 N
Explanation:
We have given the cube of iron and cube of wood both have volume [tex]V=1.65\times 10^{-4}m^3[/tex]
Density of wood [tex]\rho _{wood}=3.72\times 10^2kg/m^3[/tex]
And density of the iron [tex]\rho _{iron}=7.86\times 10^3kg/m^3[/tex]
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
We know that buoyant force is given by [tex]F=\rho Vg[/tex]
So buoyant force on the wood = [tex]3.73\times 10^2\times 1.65\times 10^{-4}\times 9.81=0.603N[/tex]
So buoyant force on iron = [tex]7.86\times 10^3\times 1.65\times 10^{-4}\times 9.81=12.722N[/tex]
